3.4 ThePoint-SlopeFormofaLine - College of the...

Section 3.4 The Point-Slope Form of a Line 293

Version: Fall 2007

3.4 The Point-Slope Form of a LineIn the last section, we developed the slope-intercept form of a line (y = mx+ b). Theslope-intercept form of a line is applicable when you’re given the slope and y-interceptof the line. However, there will be times when the y-intercept is unknown.

Suppose for example, that you are asked to find the equation of a line that passesthrough a particular point P (x0, y0) with slope = m. This situation is pictured inFigure 1.

x

y

P (x0,y0)

Q(x,y)

Figure 1. A line through (x0, y0) withslope m.

Let the point Q(x, y) be an arbitrary point on the line. We can determine the equationof the line by using the slope formula with points P and Q. Hence,

Slope = ∆y∆x

= y − y0x− x0

.

Because the slope equals m, we can set Slope = m in this last result to obtain

m = y − y0x− x0

.

If we multiply both sides of this last equation by x− x0, we get

m(x− x0) = y − y0,

or exchanging sides of this last equation,

y − y0 = m(x− x0).

This last result is the equation of the line.

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294 Chapter 3 Linear Functions

Version: Fall 2007

The Point-Slope Form of a Line. If line L passes through the point (x0, y0)and has slope m, then the equation of the line is

y − y0 = m(x− x0). (1)

This form of the equation of a line is called the point-slope form.

To use the point-slope form of a line, follow these steps.

Procedure for Using the Point-Slope Form of a Line. When given the slopeof a line and a point on the line, use the point-slope form as follows:

1. Substitute the given slope for m in the formula y − y0 = m(x− x0).2. Substitute the coordinates of the given point for x0 and y0 in the formulay − y0 = m(x− x0).

For example, if the line has slope −2 and passes through the point (3, 4), thensubstitute m = −2, x0 = 3, and y0 = 4 in the formula y − y0 = m(x − x0) toobtain

y − 4 = −2(x− 3).

I Example 2. Draw the line that passes through the point P (−3,−2) and has slopem = 1/2. Use the point-slope form to determine the equation of the line.

First, plot the point P (−3,−2), as shown in Figure 2(a). Starting from the pointP (−3,−2), move 2 units to the right and 1 unit up to the point Q(−1,−1). The linethrough the points P and Q in Figure 2(a) now has slope m = 1/2.

x

y

P (−3,−2)Q(−1,−1)

∆x=2∆y=1

x

y

R(0,−0.5)

(a) The line through P (−3,−2)with slope m = 1/2.

(b) Checking the y-intercept.

Figure 2.


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To determine the equation of the line in Figure 2(a), we will use the point-slope formof the line

y − y0 = m(x− x0). (3)

The slope of the line is m = 1/2 and the given point is P (−3,−2), so (x0, y0) =(−3,−2). In equation (3), set m = 1/2, x0 = −3, and y0 = −2, obtaining

y − (−2) = 12

(x− (−3)),

or equivalently,

y + 2 = 12

(x+ 3). (4)

This is the equation of the line in Figure 2(a).As a check, we’ve estimated the y-intercept of the line in Figure 2(b) as R(0,−0.5).

Let’s place equation (4) in slope-intercept form to determine the exact value of they-intercept. First, distribute 1/2 to get

y + 2 = 12x+ 3

2.

Subtract 2 from both sides of this last equation.

y = 12x+ 3

2− 2

Make equivalent fractions with a common denominator and simplify.

y = 12x+ 3

2− 4

2y = 1

2x− 1

2

(5)

Comparing equation (5) with y = mx+b gives us b = −1/2. This is the exact y-valueof the y-intercept. Note that this result compares exactly with the y-value of point Rin Figure 2(b). This is a bit lucky. Don’t expect to get an exact comparison everytime. However, if the comparison is not close, look for an error in your work, either inyour computations or in your graph.

Let’s look at another example.

I Example 6. Find the equation of the line passing through the points P (−3, 2)and Q(2,−1). Place your final answer in standard form.

Again, to help keep our focus, we draw the line passing through the points P (−3, 2)and Q(2,−1) in Figure 3.


Version: Fall 2007

x

y

P (−3,2)

Q(2,−1)

Figure 3. The line through pointsP (−3, 2) and Q(2,−1).

Use the slope formula to determine the slope of the line through the points P (−3, 2)and Q(2,−1).

m = ∆y∆x

= −1− 22− (−3)

= −35

We’ll use the point-slope form of the line

y − y0 = m(x− x0). (7)

Let’s use point P (−3, 2) as the given point (x0, y0). That is, (x0, y0) = (−3, 2). Sub-stitute m = −3/5, x0 = −3, and y0 = 2 in equation (7), obtaining

y − 2 = −35(x− (−3)). (8)

This is the equation of the line passing through the points P and Q.Alternatively, we could also use the point Q(2,−1) as the given point (x0, y0). That

is, (x0, y0) = (2,−1). Substitute m = −3/5, x0 = 2, and y0 = −1 in the point-slopeform (7), obtaining

y − (−1) = −35(x− 2). (9)

This too, is the equation of the line passing through the points P and Q.How can the equations (8) and (9) both be the equation of the line through P

and Q, yet look so distinctly different? Let’s place each equation in standard formAx+By = C and compare the results.

If we start with equation (8) and distribute the slope,

y − 2 = −35(x− (−3))

y − 2 = −35x− 9

5.

Multiply both sides by the common denominator 5 to clear the fractions.


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5(y − 2) = 5(−3

5x− 9

5

)5y − 10 = −3x− 9

Add 3x to both sides of the equation, then add 10 to both sides of the equation toobtain

3x+ 5y = 1. (10)

Place equation (9) in standard form in a similar manner. First, start with equation (9)and distribute the slope,

y − (−1) = −35(x− 2)

y + 1 = −35x+ 6

5.

Next, multiply both sides of this last result by 5 to clear the fractions from the equation.

5 (y + 1) = 5(−3

5x+ 6

5

)5y + 5 = −3x+ 6

Finally, add 3x to both sides of the equation, then subtract 5 from both sides of theequation to obtain

3x+ 5y = 1. (11)

Note that equation (11) is identical to equation (10). Thus, it doesn’t matter whichpoint you use in the point-slope form. Both lead to the same result.

Parallel LinesRecall that slope controls the “steepness” of a line. Consequently, if two lines areparallel, they must have the same “steepness” or slope. Let’s look at an example ofparallel lines.

I Example 12. Find the equation of the line that passes through the point P (−2, 2)that is parallel to the line passing through the points Q(−3,−1) and R(2, 1).

First, to help us stay focused, we draw the line through the points Q(−3,−1) andR(2, 1), then plot the point P (−2, 2), as shown in Figure 4(a).

We can use the slope formula to calculate the slope of the line passing through thepoints Q(−3,−1) and R(2, 1).

m = ∆y∆x

= 1− (−1)2− (−3)

= 25


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x

y

P (−2,2)

Q(−3,−1)

R(2,1)x

y

P (−2,2)

Q(−3,−1)

R(2,1)∆x=5

∆y=2

T (3,4)

(a) The line throughQ(−3,−1) and R(2, 1).

(b) The line through P (−2, 2)that is parallel to theline through Q and R.

Figure 4.

We now draw a line through the point P (−2, 2) that is parallel to the line throughthe points Q and R. Parallel lines must have the same slope, so we start at the pointP (−2, 2), “run” 5 units to the right, then “rise” 2 units up to the point T (3, 4), asshown in Figure 4(b).

We seek the equation of the line through the points P and T . We’ll use the point-slope form of the line

y − y0 = m(x− x0). (13)

We’ll use the point P (−2, 2) as the given point (x0, y0). That is, (x0, y0) = (−2, 2).The line through P has slope 2/5. Substitute m = 2/5, x0 = −2, and y0 = 2 inequation (13) to obtain

y − 2 = 25(x− (−2)). (14)

Let’s place the equation (14) in standard form. Distribute the slope, then clearfractions by multiplying both sides of the resulting equation by 5.

y − 2 = 25x+ 4

5

5(y − 2) = 5(

25x+ 4

5

)5y − 10 = 2x+ 4

Finally, subtract 5y from both sides of the last equation, then subtract 4 from bothsides of the equation, obtaining

−14 = 2x− 5y,

or equivalently,


Version: Fall 2007

2x− 5y = −14.

This is the standard form of the equation of the line passing through the point P andparallel to the line passing through the points Q and R.

Perpendicular LinesSuppose that two lines L1 and L2 have slopes m1 and m2, respectively. Recall (see thesection on Slope) that if L1 and L2 are perpendicular, then the product of their slopesis m1m2 = −1. Alternatively, the slope of the first line is the negative reciprocal ofthe second line, and vice-versa; i.e., m1 = −1/m2 and m2 = −1/m1. Let’s look at anexample of perpendicular lines.

I Example 15. Find the equation of the line passing through the point P (−4,−4)that is perpendicular to the line 4x+ 3y = 12.

It will help our focus if we draw the given line 4x + 3y = 12. The easiest way toplot a line in standard form Ax+By = C is to find the x- and y-intercepts.

4x+ 3y = 12 4x+ 3y = 12

4x+ 3(0) = 12 4(0) + 3y = 12

4x = 12 3y = 12

x = 3 y = 4

Plot the x- and y-intercepts R(3, 0) and S(0, 4) as shown in Figure 5(a). The linethrough points R and S is the graph of the equation 4x+ 3y = 12.

x

y

R(3,0)

S(0,4)

x

y

P (−4,−4)

Q(0,−1)

∆x=4

∆y=3m=−4/3

(a) Plot the x andy-intercepts of 4x + 3y = 12.

(b) A line through P (−4,−4) thatis perpendicular to 4x + 3y = 12.

Figure 5.

Next, determine the slope of the line 4x + 3y = 12 by placing this equation inslope-intercept form (i.e., solve the equation 4x+ 3y = 12 for y).2


Version: Fall 2007

4x+ 3y = 123y = −4x+ 12

y = −43x+ 4

If two lines are perpendicular, then their slopes are negative reciprocals of oneanother. Therefore, the slope of the line that is perpendicular to the line 4x+ 3y = 12(which has slope −4/3) is m = 3/4. Our second line must pass through the pointP (−4,−4). To draw this second line, first plot the point P (−4,−4), then move 4 unitsto the right and 3 units upward to the point Q(0,−1), as shown in Figure 5(b). Theline through the points P and Q is perpendicular to the line 4x+ 3y = 12.3

To determine the equation of the line through the points P and Q, we will use thepoint-slope form of the line, namely

y − y0 = m(x− x0). (16)

The slope of the line through points P and Q ism = 3/4. If we use the point P (−4,−4),then (x0, y0) = (−4,−4). Set m = 3/4, x0 = −4, and y0 = −4 in equation (16),obtaining

y − (−4) = 34

(x− (−4)),

or equivalently,

y + 4 = 34

(x+ 4). (17)

Alternatively, we could use the slope-intercept form of the line. We know that theline through points P and Q in Figure 5(b) crosses the y-axis at Q(0,−1). So, withslope m = 3/4 and y-coordinate of the y-intercept b = −1, the slope-intercept formy = mx+ b becomes

y = 34x− 1.

On the other hand, if we solve equation (17) for y,

y + 4 = 34

(x+ 4)

y + 4 = 34x+ 3

y = 34x− 1.

(18)

Note that this is identical to the result found using the slope-intercept form above.

If you also remember that the slope of Ax+ By = C is m = −A/B, then the slope of 4x+ 3y = 12 is2

m = −A/B = −4/3.It’s a good exercise to measure the angle between the two lines with a protractor. If the angle measures3

90 degrees, then you know the lines are truly perpendicular.


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It is comforting to note that the two forms (point-slope and slope-intercept) givethe same result, but how do we determine the most efficient form to use for a particularproblem? Here’s a good hint.

Determining the Form of the Line to Use. Here is some sound advice whenyou are trying to determine whether to use the slope-intercept form or the point-slope form of a line.

• If you are given the slope and the y-intercept, use the slope-intercept formy = mx+ b.

• If you are given a point (other than the y-intercept) and the slope, use thepoint-slope form y − y0 = m(x− x0).

Applications of Linear FunctionsIn this section we will look at some applications of linear functions. We begin bydeveloping a function relating Fahrenheit and Celsius temperature.

I Example 19. Water freezes at 32◦ F and 0◦C. Water boils at 212◦ F and 100◦C.F and C are abbreviations for Fahrenheit and Celsius temperature scales, respectively.Assuming a linear relationship, develop a model relating Fahrenheit and Celsius tem-perature.

First, to help keep our focus, we set up a coordinate system on a sheet of graphpaper. In Figure 6, we’ve decided to make the Celsius temperature the dependentvariable and have assigned the Celsius temperature to the vertical axis. Similarly,we’ve declared the Fahrenheit temperature the independent variable and assigned it tothe horizontal axis.4

Interpret the given data:

• Water freezes at 32◦ F and 0◦C. This gives us the point (F,C) = (32, 0), which weplot in Figure 6.

• Water boils at 212◦ F and 100◦C. This gives us the point (F,C) = (212, 100), whichwe plot in Figure 6.

Now we are on familiar ground. We want to find the equation of the line throughthese two points, which is the same type of problem we tackled in Example 6. First,use the points (32, 0) and (212, 100) to determine the slope of the line.

m = ∆C∆F

= 100− 0212− 32

= 100180

= 59.

We will now use the point-slope form of the line, y− y0 = m(x−x0) with m = 5/9 and(x0, y0) = (32, 0). Substitute m = 5/9, x0 = 32, and y0 = 0 in y − y0 = m(x − x0) toobtain

We could easily reverse these assignments, placing the Fahrenheit temperature on the vertical axis and4

the Celsius temperature on the horizontal axis.


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F250

C120

(32, 0)

(212, 100)

Figure 6. Plotting Celsius temperature versus Fahren-heit temperature.

y − 0 = 59(x− 32). (20)

However, our dependent axis is labeled C, not y, and our independent axis is labeledF , not x. So, we must replace y and x in equation (20) with C and F , respectively,obtaining

C = 59(F − 32). (21)

This result in equation (21) expresses the Celsius temperature as a function of theFahrenheit temperature. Alternatively, we could also use function notation and write

C(F ) = 59(F − 32).

Suppose that we know that the Fahrenheit temperature outside is 80◦ F and wewish to express this using the Celsius scale. To do so, we simply evaluate C(80), as in

C(80) = 59(80− 32) ≈ 26.6.

Hence, the Celsius temperature is approximately 26.6◦C.On the other hand, suppose that we know the Celsius temperature on a metal roof

is 80◦C and we wish to find the Fahrenheit temperature. To do so, we need to solve

C(F ) = 80

for F , or equivalently,59(F − 32) = 80.

Multiply both sides by 9 to obtain

5(F − 32) = 720,


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then divide both sides of the result by 5 to obtain

F − 32 = 144.

Adding 32 to both sides of this last result produces the Fahrenheit temperature F =176◦ F. Wow, that’s hot!

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